JEE MAIN - Mathematics (2023 - 1st February Evening Shift - No. 2)
Let $$P(S)$$ denote the power set of $$S=\{1,2,3, \ldots ., 10\}$$. Define the relations $$R_{1}$$ and $$R_{2}$$ on $$P(S)$$ as $$\mathrm{AR}_{1} \mathrm{~B}$$ if $$\left(\mathrm{A} \cap \mathrm{B}^{\mathrm{c}}\right) \cup\left(\mathrm{B} \cap \mathrm{A}^{\mathrm{c}}\right)=\emptyset$$ and $$\mathrm{AR}_{2} \mathrm{~B}$$ if $$\mathrm{A} \cup \mathrm{B}^{\mathrm{c}}=\mathrm{B} \cup \mathrm{A}^{\mathrm{c}}, \forall \mathrm{A}, \mathrm{B} \in \mathrm{P}(\mathrm{S})$$. Then :
only $$R_{2}$$ is an equivalence relation
both $$R_{1}$$ and $$R_{2}$$ are not equivalence relations
both $$R_{1}$$ and $$R_{2}$$ are equivalence relations
only $$R_{1}$$ is an equivalence relation
Explanation
$\begin{aligned} & \mathrm{S}=\{1,2,3, \ldots \ldots 10\} \\\\ & \mathrm{P}(\mathrm{S})=\text { power set of } \mathrm{S} \\\\ & \mathrm{AR}_1 \mathrm{B} \Rightarrow(\mathrm{A} \cap \overline{\mathrm{B}}) \cup(\overline{\mathrm{A}} \cap \mathrm{B})=\phi \\\\ & \mathrm{R}_1 \text { is reflexive, symmetric } \\\\ & \text { For transitive } \\\\ & (\mathrm{A} \cap \overline{\mathrm{B}}) \cup(\overline{\mathrm{A}} \cap \mathrm{B})=\phi ;\{\mathrm{a}\}=\phi=\{\mathrm{b}\} \therefore \mathrm{A}=\mathrm{B} \\\\ & (\mathrm{B} \cap \overline{\mathrm{C}}) \cup(\overline{\mathrm{B}} \cap \mathrm{C})=\phi \therefore \mathrm{B}=\mathrm{C} \\\\ & \therefore \mathrm{A}=\mathrm{C} \text { equivalence. }\end{aligned}$
_1st_February_Evening_Shift_en_2_1.png)
$$ \mathrm{R}_2 \equiv \mathrm{A} \cup \overline{\mathrm{B}}=\overline{\mathrm{A}} \cup \mathrm{B} $$
$\mathrm{R}_2 \rightarrow$ Reflexive, symmetric
For transitive :
_1st_February_Evening_Shift_en_2_2.png)
$\begin{aligned} & \mathrm{A} \cup \overline{\mathrm{B}}=\overline{\mathrm{A}} \cup \mathrm{B} \Rightarrow\{\mathrm{a}, \mathrm{c}, \mathrm{d}\}=\{\mathrm{b}, \mathrm{c}, \mathrm{d}\} \\\\ & \{\mathrm{a}\}=\{\mathrm{b}\} \therefore \mathrm{A}=\mathrm{B} \\\\ & \mathrm{B} \cup \overline{\mathrm{C}}=\overline{\mathrm{B}} \cup \mathrm{C} \Rightarrow \mathrm{B}=\mathrm{C} \\\\ & \therefore \mathrm{A}=\mathrm{C} \quad \therefore \mathrm{A} \cup \overline{\mathrm{C}}=\overline{\mathrm{A}} \cup \mathrm{C} \therefore \text { Equivalence }\end{aligned}$
$$ \mathrm{R}_2 \equiv \mathrm{A} \cup \overline{\mathrm{B}}=\overline{\mathrm{A}} \cup \mathrm{B} $$
$\mathrm{R}_2 \rightarrow$ Reflexive, symmetric
For transitive :
$\begin{aligned} & \mathrm{A} \cup \overline{\mathrm{B}}=\overline{\mathrm{A}} \cup \mathrm{B} \Rightarrow\{\mathrm{a}, \mathrm{c}, \mathrm{d}\}=\{\mathrm{b}, \mathrm{c}, \mathrm{d}\} \\\\ & \{\mathrm{a}\}=\{\mathrm{b}\} \therefore \mathrm{A}=\mathrm{B} \\\\ & \mathrm{B} \cup \overline{\mathrm{C}}=\overline{\mathrm{B}} \cup \mathrm{C} \Rightarrow \mathrm{B}=\mathrm{C} \\\\ & \therefore \mathrm{A}=\mathrm{C} \quad \therefore \mathrm{A} \cup \overline{\mathrm{C}}=\overline{\mathrm{A}} \cup \mathrm{C} \therefore \text { Equivalence }\end{aligned}$
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